Methods for Finding Cartesian and Polar Equations from Curves
In many cases, we may have a pair of parametric equations but find that it is simpler to draw a curve if the equation involves only two variables, such as [latex]x[/latex] and [latex]y[/latex]. Eliminating the parameter is a method that may make graphing some curves easier. However, if we are concerned with the mapping of the equation according to time, then it will be necessary to indicate the orientation of the curve as well. There are various methods for eliminating the parameter [latex]t[/latex] from a set of parametric equations; not every method works for every type of equation. Here we will review the methods for the most common types of equations.
Eliminating the Parameter from Polynomial, Exponential, and Logarithmic Equations
For polynomial, exponential, or logarithmic equations expressed as two parametric equations, we choose the equation that is most easily manipulated and solve for [latex]t[/latex]. We substitute the resulting expression for [latex]t[/latex] into the second equation. This gives one equation in [latex]x[/latex] and [latex]y[/latex].
Given [latex]x\left(t\right)={t}^{2}+1[/latex] and [latex]y\left(t\right)=2+t[/latex], eliminate the parameter, and write the parametric equations as a Cartesian equation.
We will begin with the equation for [latex]y[/latex] because the linear equation is easier to solve for [latex]t[/latex].
[latex]\begin{gathered}y=2+t \\ y - 2=t\end{gathered}[/latex]
Next, substitute [latex]y - 2[/latex] for [latex]t[/latex] in [latex]x\left(t\right)[/latex].
[latex]\begin{align}&x={t}^{2}+1 \\ &x={\left(y - 2\right)}^{2}+1&& \text{Substitute the expression for }t\text{ into }x. \\ &x={y}^{2}-4y+4+1 \\ &x={y}^{2}-4y+5 \\ &x={y}^{2}-4y+5 \end{align}[/latex]
The Cartesian form is [latex]x={y}^{2}-4y+5[/latex].
Analysis of the Solution
This is an equation for a parabola in which, in rectangular terms, [latex]x[/latex] is dependent on [latex]y[/latex]. From the curve’s vertex at [latex]\left(1,2\right)[/latex], the graph sweeps out to the right. In this section, we consider sets of equations given by the functions [latex]x\left(t\right)[/latex] and [latex]y\left(t\right)[/latex], where [latex]t[/latex] is the independent variable of time. Notice, both [latex]x[/latex] and [latex]y[/latex] are functions of time; so in general [latex]y[/latex] is not a function of [latex]x[/latex].
Given the equations below, eliminate the parameter and write as a rectangular equation for [latex]y[/latex] as a function of [latex]x[/latex].
The Cartesian form is [latex]y=\frac{3}{x}[/latex].
Analysis of the Solution
The domain is restricted to [latex]t>0[/latex]. The Cartesian equation, [latex]y=\frac{3}{x}[/latex] has only one restriction on the domain, [latex]x\ne 0[/latex].
Eliminate the parameter and write as a Cartesian equation: [latex]x\left(t\right)=\sqrt{t}+2[/latex] and [latex]y\left(t\right)=\mathrm{log}\left(t\right)[/latex].
The Cartesian form is [latex]y=\mathrm{log}{\left(x - 2\right)}^{2}[/latex].
Analysis of the Solution
To be sure that the parametric equations are equivalent to the Cartesian equation, check the domains. The parametric equations restrict the domain on [latex]x=\sqrt{t}+2[/latex] to [latex]t>0[/latex]; we restrict the domain on [latex]x[/latex] to [latex]x>2[/latex]. The domain for the parametric equation [latex]y=\mathrm{log}\left(t\right)[/latex] is restricted to [latex]t>0[/latex]; we limit the domain on [latex]y=\mathrm{log}{\left(x - 2\right)}^{2}[/latex] to [latex]x>2[/latex].
Eliminate the parameter and write as a rectangular equation.
